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A322323 Self-stuffable numbers (see the Comments section for definition). 3
20, 22, 120, 126, 300, 2200, 4000, 50000, 150000, 202020, 231000, 240000, 600000, 1600000, 2500000, 7000000, 12500000, 14300000, 40400000, 44000000, 80000000, 101250000, 102600000, 108000000, 112500000, 120120120, 126000000, 131400000, 144000000, 180000000, 225000000, 300300300, 312300000 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

See A322322 for the definition of a stuffable number in general.

The sequence is infinite because it contains the numbers 202020...20 (that is, k copies of 20) for any odd k. - Hans Havermann, Dec 03 2018

Do we have enough terms now to make a guess about exactly what numbers are in this sequence? - N. J. A. Sloane, Dec 06 2018

From M. F. Hasler, Dec 07 2018: (Start)

Def. 1: Let E(n) = sumdigits(n) - #digits(n) = A007953(n) - A055642(n). We call N > 0 admissible iff E(N) = A010879(N), the last digit of N.

Remark: Any term of this sequence must be admissible. If N is admissible then any permutation of N's digits with fixed last digit (and no leading 0) is again admissible. Also, the last digit of N can be replaced by any arbitrary digit.

Prop. 1: If E(n) >= 0, then N = n*10^E(n) is admissible. (Obvious from definitions.)

Remark: Most terms of this sequence are of that form, with some n not a multiple of 10. Are there any terms not of this form, beyond a(2) = 22 and a(4) = 126?

Prop. 2: If N is admissible and a multiple of 10, then the concatenation of N and any other admissible number M is again admissible. In particular, M can be 1 or any number of concatenations of N. Instead of N one can take an arbitrary number of admissible multiples of 10, (N1, ..., Nk).

Def. 2: When N is admissible, let S(N) be the result of N stuffed with itself, i.e., after each nonzero digit d[i] of N except for the last digit, we insert d[i] digits of N, taken one after the other starting with d[1], d[2], etc.: after d[1] are inserted d[1], ..., d[d[1]]; after d[2], if nonzero, are inserted d[d[1]+1], ..., d[d[1]+d[2]], etc.

Remark: The last digit inserted, just before the last digit d[L], is also d[d[1]+...+d[L-1]] = d[L]. Thus, the result S(N) has both its first and its last digit occurring twice in a row.

Def. 3: We say N is self-stuffable iff N is admissible and N divides S(N).

Example: For 0 < k < 9, N = (k+1)*10^k is self-stuffable, thus in this sequence.

Prop. 3: If N and M are admissible and N = 0 (mod 10), then S(concat(N,M)) = concat(S(N),S(M)). The same is true with any number of copies of N.

Lemma: If N is admissible, then S(N*10^E(N)) = S(N)*10^E(N). (Similar to Prop. 1.)

Corollary: If M is admissible, then so are N = M*10^E(M) and T = concat(N,N,M) = (100^L + 10^L + 1)*M with L = #digits(N), and S(T) = (10^(4L) + 100^L + 1)*S(M) = (100^L - 10^L + 1)*T*S(M)/M. Thus, if N is self-stuffable, then so is T.

More generally, Sum_{k=0..m} x^(2k) = Sum_{k=0..m} (-x)^k * Sum_{k=0..m} x^k for any even m, and therefore (with x = 10^L):

Theorem: If M is self-stuffable, then so is M_m := Sum_{k=0..2m} 10^(k*L)*M with L = #digits(M) + E(M) for all m >= 0. (This is the concatenation of 2m copies of M*10^E(M) and then M.)

Example: For M = 20, E(M) = 0, this is Hans Havermann's comment above. For M = 22, E(M) = 2, it asserts that 22002200...22 (any odd number of '22's) is in the sequence; similarly for M = 126, E(M) = 6, not only 126*10^6 but also 126*(10^18 + 10^9 + 1) etc. are self-stuffable.

We may call a self-stuffable (or an admissible) number primitive if it is not of the form M_m with m > 0. The smallest integer M > 0 such that a(n) = M_m or a(n) = (M*10^E(M))_m will be called the root of a(n). It is easy to see that any term has a root which is not a multiple of 10.

The sequence can be computed very efficiently as follows: Consider numbers M > 0 that are not a multiple of 10. If E(M) > 0 and N | S(N) for N = M*10^E(M), then all {N_m; m>=0} are in the sequence. If in addition to this M is admissible and M | S(M), then all {M_m; m>=0} are also in the sequence. This yields all terms. (End)

The list of roots is now given in A322002. - M. F. Hasler, Dec 09 2018

For nonnegative k, 10^9*(103*10^k+(10^k-1)/9)+8*10^6 will always be self-stuffable, thus ensuring that there will always be at least one d-digit solution when d > 11. - Hans Havermann, Dec 12 2018

LINKS

John Mason, Table of n, a(n) for n = 1..145 (Terms 1..41 from Hans Havermann)

EXAMPLE

126 is in the sequence as 126 will be opened as 1.2..6 and the dots can be filled by 126 itself (as 112266 is divisible by 126: 112266/126 = 891).

1200 isn't a term because 1.2..0 hasn't enough empty places to put the four digits of 1200 (the last 0 isn't concatenated to the right, else 1200 would be a term). - David A. Corneth, Dec 09 2018

MATHEMATICA

Select[Range[10, 10^6], Block[{n = #1, d = #2, w = #3, s = Position[#3, {}][[All, 1]]}, If[Length@ d == Length@ s, Mod[FromDigits@ ReplacePart[w, Array[s[[#]] -> d[[#]] &, Length@ s]], n] == 0, False]] & @@ {#1, #2, Drop[Flatten[Map[Prepend[ConstantArray[{}, #], #] &, #2], 1], -Last@ #2]} & @@ {#, IntegerDigits[#]} &] (* Michael De Vlieger, Dec 08 2018 *)

PROG

(PARI) is(n, d=digits(n), i=0)={#d==vecsum(d)-d[#d] && fromdigits(concat(vector(#d, j, if( d[j]&& j<#d, vector(d[j]+1, k, d[if(k>1, i+=1, j)]), d[j]))))%n==0} \\ M. F. Hasler, Dec 07 2018

CROSSREFS

A322322 gives the definition of a stuffable number.

A322002 lists the roots which produce all terms of this sequence.

Cf. A007953, A010879, A055642.

Sequence in context: A163630 A259735 A033906 * A139357 A104059 A104057

Adjacent sequences:  A322320 A322321 A322322 * A322324 A322325 A322326

KEYWORD

base,nonn,changed

AUTHOR

Eric Angelini and Jean-Marc Falcoz, Dec 03 2018

STATUS

approved

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Last modified January 20 13:11 EST 2019. Contains 319332 sequences. (Running on oeis4.)